# I Polynomial of finite degree actually infinite degree?

#### Mr Davis 97

$1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k$.

Isn't this a contradiction since the LHS has degree $2$ while the RHS has infinite degree?

#### fresh_42

Mentor
2018 Award
$1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k$.

Isn't this a contradiction since the LHS has degree $2$ while the RHS has infinite degree?
Wherever your polynomials live in, $\mathbb{F}(x)\; , \;\mathbb{F}[[x]]\,,$ it only says that this field doesn't have a proper evaluation function given by the degree, or worse, that it does not included infinite (quotients of) polynomials.

#### Math_QED

Homework Helper
$1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k$.

Isn't this a contradiction since the LHS has degree $2$ while the RHS has infinite degree?
How do you define the degree of a polynomial in $\mathbb{F}[[X]]$?

#### Mr Davis 97

How do you define the degree of a polynomial in $\mathbb{F}[[X]]$?
To me the degree of a polynomial is the highest degree of its monomials with non-zero coefficients. I don't really know how to define degree precisely in the infinite case.

Actually, now that I expand the product, we get that all of the terms of degree higher than 2 cancel out, so I guess there is no problem.

But I do have another, slightly related question. We have that $1+x+x^2 = \dfrac{1-x^3}{1-x}$. But the LHS can be evaluated at x=1 while the RHS cannot. How are they equal then?

#### fresh_42

Mentor
2018 Award
To me the degree of a polynomial is the highest degree of its monomials with non-zero coefficients. I don't really know how to define degree precisely in the infinite case.

Actually, now that I expand the product, we get that all of the terms of degree higher than 2 cancel out, so I guess there is no problem.

But I do have another, slightly related question. We have that $1+x+x^2 = \dfrac{1-x^3}{1-x}$. But the LHS can be evaluated at x=1 while the RHS cannot. How are they equal then?
They are not equal. That's why I said: wherever your polynomials live. A rational polynomial and a rational polynomial function are not the same thing. But as just functions, they are not the same, because the quotient does not have the same range as the polynomial. The graph of the quotient has a hole!

#### fresh_42

Mentor
2018 Award
Let me ask something in return: Why is $x^3-x = 0\,.$

#### fbs7

$1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k$.

Isn't this a contradiction since the LHS has degree $2$ while the RHS has infinite degree?
Hmm... aren't they the same thing? Assuming the series converge,

(1-x3)(1+x+x2+x3+x4+x5+...)
= (1+x+x2+x3+x4+x5+...)
- x3(1+x+x2+x3+x4+x5+...)
= (1+x+x2)+(x3+x4+x5+...)-(x3+x4+x5+...)
= 1+x+x2

The OP is similar to asking why the LHS is a 3-degree polymial and the RHS is a 10-degree polynomial... I don't think the RHS is a 10-degree polynomial:

1+x+x2 = 1+x+x2 + x10 - x10

#### Math_QED

Homework Helper
Hmm... aren't they the same thing? Assuming the series converge,

(1-x3)(1+x+x2+x3+x4+x5+...)
= (1+x+x2+x3+x4+x5+...)
- x3(1+x+x2+x3+x4+x5+...)
= (1+x+x2)+(x3+x4+x5+...)-(x3+x4+x5+...)
= 1+x+x2

The OP is similar to asking why the LHS is a 3-degree polymial and the RHS is a 10-degree polynomial... I don't think the RHS is a 10-degree polynomial:

1+x+x2 = 1+x+x2 + x10 - x10
These are formal power series. This is algebra, no analysis. Nothing converges here, these are just formal expressions that can be manipulated.

#### fbs7

Hmm... that's surprising...

Say that x = 2; then 1-x3 = -7, while (1+x+x2+...) = ∞, therefore (1-x3)*(1+x+x2+...) = -7 . ∞ = -∞

Meanwhile (1+x+x2+x3+x4+...) - (x3+x4+....) = ∞ - ∞ = indeterminate

I was trying to avoid assuming true a step that would be indeterminate if the series diverged, but if that's not necessary, then I stand corrected.

#### fresh_42

Mentor
2018 Award
Hmm... that's surprising...

Say that x = 2; then 1-x3 = -7, while (1+x+x2+...) = ∞, therefore (1-x3)*(1+x+x2+...) = -7 . ∞ = -∞

Meanwhile (1+x+x2+x3+x4+...) - (x3+x4+....) = ∞ - ∞ = indeterminate

I was trying to avoid assuming true a step that would be indeterminate if the series diverged, but if that's not necessary, then I stand corrected.
You first have to say what those polynomials mean:
• formal power series
• and if, invertible or not
• real functions
Whether they allow the definition of a degree, a substitution map like $x=2$, infinite length or whatever depends on this category. $1+x+x^2+\ldots$ is without meaning if we do not attach one. As this meaning wasn't given, all we have are sequences:
\begin{align*}
1-x^3 &\triangleq (1,0,0,-1,0,0,\ldots) \\
1+x+x^2+\cdot & \triangleq (1,1,1,\ldots)
\end{align*}
and all they bring is a linear structure (addition and scalar multiplication) and perhaps a piecewise multiplication, which is not the usual polynomial multiplication. So neither polynomial multiplication, nor inversion, nor substitution, nor a degree, nor topological methods can be applied without further specifications.

It is the old fraud: start with something familiar, transfer it in a hidden world and draw false conclusions in the first environment.
Hint: This method is far better paid in Las Vegas, where it is called performances of magicians.

#### fbs7

Wow, I'm in deep and unchartered waters now! I thought that

$1+x+x^2+x^3+...$

always meant a limit

$\lim_ {n \rightarrow +\infty} \sum_{i=0}^{n} x^i$

but it may mean other things! I gotta think about that over a hotdog. Living and learning!

Code:
int main(void)
{
Dude fbs;
fbs.ImproveMathSkill(0.00001,"%");
fbs.GetFood(new HotDog());
}

• DarMM

#### jbriggs444

Wow, I'm in deep and unchartered waters now! I thought that

$1+x+x^2+x^3+...$

always meant a limit
If you interpret the notation as a possibly divergent series then one can explore ways in which to evaluate the "sum" of a divergent series. https://en.wikipedia.org/wiki/Divergent_series

If you interpret the notation as a polynomial then you the advice of @fresh_42 is appropriate. One can distinguish between...

A formal polynomial: One can manipulate add, subtract and multiply formal polynomials by simply looking at the list of coefficients on each polynomial and coming up with a list of coefficients for the sum, difference or product. When looked at this way, there is no need to consider evaluating the polynomial or whether evaluation is even possible.

A polynomial function: One can treat a polynomial as a function produces a result, f(x) for each input, x. One can add, subtract or multiply polynomial functions by adding, subtracting or multiplying the evaluated results. When looked at this way, is is crucial to have a way of evaluating the polynomial for every possible input be aware of the input ranges for which evaluation is possible.

A polynomial function evaluated for a particular input. Again, one can add, subtract, multiply or divide. But one only needs to consider the selected input.

One thing that I find interesting is to contemplate polynomials over the boolean field GF(2) which contains just the integers 0 and 1. There are only four polynomial functions over this field: The constant 0 function, the constant 1 function, the identity function and the inverse function. But there are eight formal polynomials of degree 2 or less: x^2 + x + 1, x^2 + x, x^2 + 1, x^2, x + 1, x, 1 and 0.

#### fresh_42

Mentor
2018 Award
I think it is a widespread phenomenon at all levels to write down an expression and completely neglect the context, the definition or any other hint on how it should be interpreted. It is often clear by the circumstances, but as soon as it is written down, circumstances will be decoupled. We see this here quite often: members write down their problem and totally forget, that we readers are outstanders with respect to their circumstances. That's why I personally find section 2 of our template far more important than section 3.

I often try to fight these neglencies Don Quixote like as I think it is necessary to get used to a proper writing style, and actually do define what is meant - always. One of my famous objections is, if someone says $2+2=4$ or something. I'm literally forced to answer with a 'No'. If nothing else is said, we are in the naturals, the integers, the rationals, the reals or none of them? This might be nitpicking, but it is the mean seed of all later inaccuracies.

#### WWGD

Gold Member
I think the factorization doesn't go through$h: 1-x^3 =(1-x) (1+x+x^2)$ so, if $x\neq 1$ then the ratio equals 1+x+x^2, but the main point remains. Sorry, please delete post.

#### FactChecker

Gold Member
2018 Award
More simply, you should consider that $\frac {1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $0 \lt x \lt 1$. Would you say that the right-hand side is of infinite degree, when you know that the limit is a very smooth, simple, function? As @Math_QED indicated, that would be confusing algebra with analysis.

#### fbs7

Hmm... I'm kind of skimming on the almost-understanding this... So, let me try to organize my thoughts on this. This is a valid deduction

[a1] x=2 & y=x
[a2] y=2

The deduction below is false, of course.

[b1] x=2 & y=x
[b2] x=2 & y=x.(1-x)/(1-x)
[b3] x=2 & y=x.(1-x).(1+x+x2+x3+...) & 0<x<1
[b4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...)) & 0<x<1
[b5] x=2 & y=x & 0<x<1
[b6] y=x & ( x=2 & 0<x<1 )

This leads to a contradiction, so it contains a mistake. I thought that the contradiction comes from a mistake on deducing [b3] from [b2], and that mistake is that I can't manipulate an infinite series when that series diverges. But, if I understood it right up so far:

(a) in Analysis fbs7 is correct, the contradiction arises because you cannot manipulate an infinite series if that series diverges
(b) in Algebra fbs7 is incorrect, and the contradiction arises because the clause 0<x<1 is artificial and not necessary, and without that clause this would be a perfectly fine deduction:

[c1] x=2 & y=x
[c2] x=2 & y=x.(1-x)/(1-x)
[c3] x=2 & y=x.(1-x).(1+x+x2+x3+...)
[c4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[c5] x=2 & y=x
[c6] y=2

So, tell me this... if [c6] is a valid deduction in Algebra, then is this also a valid deduction in Algebra?

[d1] x=2 & y=x
[d2] x=2 & y=x + log(0) - log(0)
[d3] x=2 & y=x
[d4] y=2

I can't believe you can deduce [d3] from [d2] above, because if you could, then you could also deduce

[e1] x=2 & y=x
[e2] x=2 & y=x + log(0) - log(0)
[e3] x=2 & y=x + log(2.0) - log(0)
[e4] x=2 & y=x + log(0) + log(0) - log(0)
[e5] x=2 & y=x + log(0)
[e6] y=2 + log(0)
[e7] y=log(102.0)
[e8] y=log(0)

So how in the world in Algebra I can deduce [c5] from [c4], but I cannot deduce [d3] from [d2]?

#### fbs7

A variant of the same theme... if in Algebra I can deduce

[c4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[c5] x=2 & y=x
[c6] y=2

then why I can't deduce

[cc4] x=1 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[cc5] x=1
[cc6] from [cc4], y=x.((1+x+x2+x3+...)-1*(x+x2+x3+...))
[cc7] from [cc5], y=x.((1+x+x2+x3+...)-x*(x+x2+x3+...))
[cc8] y=x.((1+x+x2+x3+...)-(x2+x3+x4+...))
[cc9] y=x.(1+x)
[cc10] y=2

The only difference between [c4] and [cc4] is that x=2 in the first, and x=1 in the second, and in the first I'm told that's valid, while the second it's certainly a mistake, and the only thing I did on [cc6] to [cc8] was to use the fact that x=1...

#### FactChecker

Gold Member
2018 Award
I don't understand b3 of post #16. In the same statement, you are saying that x=2 and 0<x<1. That is wrong. You should not apply an infinite series with an x value where it does not converge. That is meaningless.

Last edited:

#### fbs7

I don't understand b3 of post #16. In the same statement, you are saying that x=2 and 0<x<1. That is wrong. You should not apply an infinite series with an x value where it does not converge. That is meaningless.
Precisely my point. I argued that when one replaces 1/(1-x) with an infinite series, that always adds the restriction 0<x<1, so you can't do that when x=2 and will lead to all kinds of jabberwoking. Ergo can't expand in series for values the series diverges.

But earlier in the thread I was told that the restriction 0<x<1 is not needed if the series is seen as a polynomial instead of a limit, and x*(1-x)*(.. power series...) still reduces to x even if the series diverges, ie, whatever the value of x. That I'm still trying to get in between my left ear and right ear.

• FactChecker

#### FactChecker

Gold Member
2018 Award
But earlier in the thread I was told that the restriction 0<x<1 is not needed if the series is seen as a polynomial instead of a limit, and x*(1-x)*(.. power series...) still reduces to x even if the series diverges, ie, whatever the value of x. That I'm still trying to get in between my left ear and right ear.
I am also at a loss of what to think about that. But I see there are several good mathematicians who seem to accept it, so I will leave this discussion to them.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving